Mathematics

In parallelogram $ABCD$ two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP=BQ (see fig. 8.20)$. Show that:  $AP = CQ$

SOLUTION
Given $ABCD$ is parallelogram and given $BD$ is diagonal such that

$DP= BQ$

$\Rightarrow$ Now $AD \parallel BC$ [$AD$ is parallel to $BC$]

$BD$ is transversal i.e., diagonal

In $\triangle APD$ & $\triangle CQB$

$AD= CB$

$\angle ADP= \angle CBQ$  [Diagonals bisect the angles ]

$DP= BQ$

$\therefore \triangle APD \cong \triangle CQB$ [$SAS$ concurrency ]

$\therefore AP= CQ$

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Subjective Medium Published on 09th 09, 2020
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